Are you sure that statement 1 is correct?

I took this from zuleron's 198 gmat prep questions and it obviously didn't look right to me at first either.

I think a is supposed to be 7 < x < 11.

The answer is D and it makes sense.

the median=mean, therefore, x=9.

hey what is the answer.According to my calulation is is D.

yep the answer is D.

the 5 numbers can be arranged as follows
2, 7, 11,16 and x.

Stmt 1 says x is between 7 and 11.
so the arrangement becomes
2, 7, x, 11, 16.

If u look at this seq carefully , u will notice 2 and 7, 11 and 16 are evenly spaced (diff = 5).
So we can ignore them in calculating the value of x.
so x cna be calculated simply as 7+11/2. = 9.
So sufficient.

Stmt 2 says the same

Hence ans is D.

HTHelps
-V

Well, if the st 1: is considered to be 7<x>11 which boils down to x> 11 then statement I is insufficient
and the answer will be "B"

as opposed to considering 7<x<11, where answer will be D.

In consecutive nos the Mean = Median. So can one conclude that if the no are arranged in ascending order 2 7 11 16 , x has to be the middle value ie.between 7 and 11. As mentioned before in this thread the diff between 7 & 2 = 11 & 16 = 5. Also the average of 2 7 11 and 16 is 9. So x cannot be in the beginning or end of the series.


Is 1 and 2 necessary as they are stating the obvious. Please point out the error in my reasoning.

Dear kstv,
you are right and wrong.
In consecutive nos the Mean = Median but the converse does not have to be true.
If mean = median, they don't have to be consecutive numbers.
Consider set A = 1, 9, 16, 25, 29 Clearly, 16 is the median
mean = (1+9+16+25+29)/5 = 16
Thus, mean = median and the numbers are not consecutive (or the set is not evenly spaced set, to be precise). Consecutive integers is special case of evenly spaced set.

kstv wrote:In consecutive nos the Mean = Median. So can one conclude that if the no are arranged in ascending order 2 7 11 16 , x has to be the middle value ie.between 7 and 11. As mentioned before in this thread the diff between 7 & 2 = 11 & 16 = 5. Also the average of 2 7 11 and 16 is 9. So x cannot be in the beginning or end of the series.


Is 1 and 2 necessary as they are stating the obvious. Please point out the error in my reasoning.

Thanks, using this can U disprove/elaborate what I have concluded about the options not being necessary.

26) If the average of five numbers, x, 7, 2, 16, and 11 = the median, what is x?

a. 7 < x < 11
b. x is median of the five numbers

Arranging terms in ascending order, we have,
2, 7, 11, 16 x can be any number.
It is given that mean = median. Depending on the value of x,
if x < 7, median = 7
7 < x < 11, median = x
x > 11, median = 11

Stmt. I
7 < x < 11
which gives us, median = mean = x
Hence, 2+7+11+16+x = 5*x
Solving, x = 9
Suff.

Stmt. II
x is the median of the five number. Essentially, this is the same info. as given in Stmt. I
Hence, Suff.

Answer is "D"

kstv,
This is a value question, which means we must find a definite value for "x". The theory for consecutive numbers does not really come into play here.

kstv wrote:Thanks, using this can U disprove/elaborate what I have concluded about the options not being necessary.

was looking for the part which you explained

Arranging terms in ascending order, we have,
2, 7, 11, 16 x can be any number.
It is given that mean = median. Depending on the value of x,
if x < 7, median = 7
7 < x < 11, median = x
x > 11, median = 11

If x is 19 Mean = Median = 11.
so x does not necessarily lie between 7 and 11 if Mean = Median.

My doubt was whether the rule of cosecutive no applies universally to a ascending series. Thanks.

but the first stmt says that 7<x>11 so doesnt it mean that 7<11<x ? how have u written as 7<x<11 ??

even if we use 7<x>11
this means x>11
and thus arranging in ascending order,
2,7,11,x,16 or 2,7,11,16,x (depends on whether x <16)
Median is still 11
and thus (2+7+11+x+6)/5 =11
and we get a single value of x.
Thus A is sufficient
also B is sufficient on its own.
Thus D